Question

Consider the two gaseous equilibria: The values of the equilibrium constants K 1 and K 2 are related by

Answer 1

The question is missing parts. The complete question is as follows.

Consider the two gaseous equilibria involving SO2 and the corresponding equilibrium constants at 298K:

`SO_(2)_((g)) + (1)/(2)O_(2)` ⇔
`SO_(3)_((g))`;
`K_(1)`

`2SO_(3)_((g))` ⇔
`2SO_(2)_((g))+O_(2)_((g)); K_(2)`

The values of the equilibrium constants are related by:

a)
`K_(1)` =
`K_(2)`

b)
`K_(2) = K_(1)^(2)`

c)
`K_(2) = (1)/(K_(1)^(2))`

d)
`K_(2)=(1)/(K_(1))`

Answer: c)
`K_(2) = (1)/(K_(1)^(2))`

Explanation: Equilibrium constant is a value in which the rate of the reaction going towards the right is the same rate as the reaction going towards the left. It is represented by letter K and is calculated as:

`K=([products]^(n))/([reagents]^(m))`

The concentration of each product divided by the concentration of each reagent. The indices, m and n, represent the coefficient of each product and each reagent.

The equilibrium constants of each reaction are:

`SO_(2)_((g)) + (1)/(2)O_(2)` ⇔
`SO_(3)_((g))`

`K_(1)=([SO_(3)])/([SO_(2)][O_(2)]^(1/2))`

`2SO_(3)_((g))` ⇔
`2SO_(2)_((g))+O_(2)_((g))`

`K_(2)=([SO_(2)]^(2)[O_(2)])/([SO_(3)]^(2))`

Now, analysing each constant, it is easy to see that
`K_(1)` is the inverse of
`K_(2)`.

If you doubled the first reaction, it will have the same coefficients of the second reaction. Since coefficients are "transformed" in power for the constant, the relationship is:

`K_(2)=(1)/(K_(1)^(2))`